The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
Learn more about momentum:
brainly.com/question/250648
#SPJ1
Click on your picture and your mobile photos will become to appear. Select the picture you want or like by double tap....
C it takes a lot of capital to get the windmills and the fields of windmills set up
If the forces are equal, at a distance equidistant it is not possible to act a pair on the body since both torques cancel each other. Being of the same magnitude and in the opposite direction, the sum of the torques will be zero.